Идёт турнир! | В настоящий момент идёт турнир. Некоторая информация и функции сервера недоступны до его окончания.
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| imesecan | 08.апр.2011 в 17:20:49 | |
| 0 | Реплика удалена
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| imesecan | 09.апр.2011 в 10:19:41 | |
| 1 | Unlucky :)
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| imesecan | 09.апр.2011 в 21:56:40 | |
| 2 | I think you don't need reverse function. When calculating the remainders you can directly add this remainder to the beginning of resulting number:
You can use multiplier as the powers of 10, The first number is to be multiplied with. Let's say that the first modulus operation gives 2 and res is initially 0 and the number is 17
res = res + 1 * rem; ==> res= 2
then for the second remainder
res = res + 10 * rem; ==> res= 2 + 2* 10 ==> 22
then on the third step
res = res + 100 * rem; ==> 22 + 1 * 100 ==> 122
so on.
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| imesecan | 19.мар.2013 в 13:15:45 | |
| 3 | Реплика удалена
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| imesecan | 24.мар.2013 в 10:46:50 | |
| 4 | Реплика удалена
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